silverglade Posted February 17, 2012 Share Posted February 17, 2012 Hi, I am trying to echo out images on my server's directory "uploads". The images are blank and not showing up. Please help if anyone can. Thank you. here is the page as it is. http://ealike.com/simple_uploader/uploader.php and here is the code. <?php // Where the file is going to be placed $target_path = "uploads/"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } //echo out all images $dir = "uploads/"; if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "<img src='$file' alt='$file' height='200' width='200' /> " ; } closedir($dh); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/ Share on other sites More sharing options...
PaulRyan Posted February 17, 2012 Share Posted February 17, 2012 Change this: echo "<img src='$file' alt='$file' height='200' width='200' /> "; To this: echo "<img src='{$dir}{$file}' alt='{$file}' height='200' width='200' />"; Regards, PaulRyan. Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318472 Share on other sites More sharing options...
silverglade Posted February 17, 2012 Author Share Posted February 17, 2012 thank you so much! man that was driving me absolutely insane. LOL. My only other problem is that it outputs two blank images with broken link icons in them, even if there are no images in the directory. Please, any more help would be greatly appreciated. thank you. here is the problem on this page http://ealike.com/imageuploader2/uploader.php and here is the nice and better updated code with a little bit more advanced functionality. <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","100"); //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['Submit'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname="images/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; }}}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h1>File Uploaded Successfully! </h1>"; } //echo out all images $dir = "images/"; if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "<img src='{$dir}{$file}' alt='{$file}' height='200' width='200' />"; } closedir($dh); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" --> <form name="newad" method="post" enctype="multipart/form-data" action=""> <table> <tr><td><input type="file" name="image"></td></tr> <tr><td><input name="Submit" type="submit" value="Upload image"></td></tr> </table> </form> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318476 Share on other sites More sharing options...
El Chupacodra Posted February 17, 2012 Share Posted February 17, 2012 Not having gone to it all too carefully I think when you check the dir it counts all your files plus . and .. so you will always get two more files than you have "real" files in your directory. If you make it a for-loop instead and count to the number of files in the directory but set the counter to start at two I think you will get it to work. Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318483 Share on other sites More sharing options...
silverglade Posted February 17, 2012 Author Share Posted February 17, 2012 thank you. I don't know how to do that part of the script with a for loop though because I don't know what variable is being set to be started at the number 2. Could you please show me what you mean in code please? if it is not too long. thank you for helping me. this is what you would be changing. //echo out all images $dir = "images/"; if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "<img src='{$dir}{$file}' alt='{$file}' height='200' width='200' />"; } closedir($dh); } } Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318485 Share on other sites More sharing options...
silverglade Posted February 17, 2012 Author Share Posted February 17, 2012 I guess this is kind of lame, but it works. I have no idea how the directory searching stuff works in detail, so I added this in the while loop. I found that solution on the net. if($file == ".") continue; if($file == "..") continue; Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318487 Share on other sites More sharing options...
El Chupacodra Posted February 17, 2012 Share Posted February 17, 2012 I was going to suggest something like this: $nrfiles = scandir($dir); $len = count($nrfiles); $nr= $len-2; for ($i=2;$i<$len;$i++) { ...echoing your photos... } But I'm glad you found a way to work with it - plus your solution took less changes to your code so it was much better, Quote Link to comment https://forums.phpfreaks.com/topic/257213-images-not-echoing-out-from-directory/#findComment-1318496 Share on other sites More sharing options...
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